$\frac{x+y}{xy+1}$ when $-1I've been reading C.C. Pinter's A Book of Abstract Algebra, and one of the excercises there is proving that the set $\{x\in \mathbb{R}:-1<x<1\}$ forms an abelian group under the operation $x*y=\frac{x+y}{xy+1}$.
The way the question is phrased doesn't require you to show that $*$ is indeed an operation on the set (it takes it for granted), but I'm having trouble with showing that to myself.
It's well-defined, because $xy+1=0$ implies $xy=-1$, which is impossible for $x,y$ in the domain.
How do I prove its closure, though? I need to show that whenever $-1<x,y<1$, we also get that $-1<x*y<1$.
I broke it into different scenarios- $x,y>0$, $x<0$ and $y>0$, or $x,y<0$.
I left out the case when $x=0$ or $y=0$ because it's simple to check, and of course, I didn't have to include ($x>0$ and $y<0$) because $x$ and $y$ are symmetric.
How would one go on from here?
 A: Here's an after-the-fact answer to "what's 'really' going on here?": The hyperbolic tangent $\tanh:(\mathbf{R}, +) \to \bigl((-1, 1), *\bigr)$ is a group isomorphism, thanks to the identity
$$
\tanh(u + v) = \frac{\tanh u + \tanh v}{1 + \tanh u \tanh v},\qquad \text{$u$, $v$ real.}
$$
(This implicitly shows $(-1, 1)$ is closed under $*$, if you buy that $\tanh$ is bijective and a sum of real numbers is a real number.)
A: Hint: you have to show that $|x+y|<|xy+1|$. It's pretty easy to see $xy>-1$, so $|xy+1|=xy+1$. Now if $|x+y|\geq0$, then $|x+y|<|xy+1|$ is equivalent to $0<(x-1)(y-1)$. Similarly for $|x+y|<0$.
A: You have that $$-1<x<1\ \ \&\ \ -1<y<1$$ Therefore,


*

*$x<1\ \&\ y<1$ implies that $x-1<0$ and $y-1<0$. Thus $$(x-1)(y-1)>0\Rightarrow xy-x-y+1>0\Rightarrow \frac{x+y}{1+xy}<1.$$ 

*$x>-1\ \&\ y>-1$ implies that $x+1>0$ and $y+1>0$. Thus $$(x+1)(y+1)>0\Rightarrow xy+x+y+1>0\Rightarrow x+y>-(1+xy)\Rightarrow\frac{x+y}{1+xy}>-1.$$


and you obtain the desired result. (Notice that $xy+1>0$).
A: There's a nice shortcut if you note that $z\in(-1,1)\iff1-z^2>0$. Then:
$$1-(x*y)^2=1-\left(\frac{x+y}{1+xy}\right)^2=\frac{(1+xy)^2-(x+y)^2}{(1+xy)^2}=\frac{(1-x^2)(1-y^2)}{(1+xy)^2}$$
and the result drops out quite nicely from here.
